Question: Simplify the following expression: $y = \dfrac{-5x^2- 2x+16}{-5x + 8}$
Explanation: First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-5)}{(16)} &=& -80 \\ {a} + {b} &=& &=& {-2} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-80$ and add them together. Remember, since $-80$ is negative, one of the factors must be negative. The factors that add up to ${-2}$ will be your ${a}$ and ${b}$ When ${a}$ is ${8}$ and ${b}$ is ${-10}$ $ \begin{eqnarray} {ab} &=& ({8})({-10}) &=& -80 \\ {a} + {b} &=& {8} + {-10} &=& -2 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-5}x^2 +{8}x) + ({-10}x +{16}) $ Factor out the common factors: $ x(-5x + 8) + 2(-5x + 8)$ Now factor out $(-5x + 8)$ $ (-5x + 8)(x + 2)$ The original expression can therefore be written: $ \dfrac{(-5x + 8)(x + 2)}{-5x + 8}$ We are dividing by $-5x + 8$ , so $-5x + 8 \neq 0$ Therefore, $x \neq \frac{8}{5}$ This leaves us with $x + 2; x \neq \frac{8}{5}$.